JEE MAIN - Physics (2018 - 15th April Morning Slot - No. 3)
One mole of an ideal monoatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature, $${27^ \circ }C.$$ The work done on the gas will be :
$$300$$ $$R$$
$$300$$ $$R$$ $$ln$$ $$6$$
$$300$$ $$R$$ $$ln$$ $$2$$
$$300$$ $$R$$ $$ln$$ $$7$$
Explanation
We know,
Work done on gas = nRT $$\ell $$n $$\left( {{{{P_f}} \over {{P_i}}}} \right)$$
Given Pf = 2Pi
T = 27 + 273 = 300 K
and for monoatomic gas, n = 1.
$$\therefore\,\,\,\,$$ Work done = 1 $$ \times $$R $$ \times $$300 $$\ell $$n(2)
= 300 R $$\ell $$n(2)
Work done on gas = nRT $$\ell $$n $$\left( {{{{P_f}} \over {{P_i}}}} \right)$$
Given Pf = 2Pi
T = 27 + 273 = 300 K
and for monoatomic gas, n = 1.
$$\therefore\,\,\,\,$$ Work done = 1 $$ \times $$R $$ \times $$300 $$\ell $$n(2)
= 300 R $$\ell $$n(2)
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